If A is an invertible matrix of order 3, then
Match List-I with List-II
|
List-I |
List-II |
|
(A) $|adj\,A|$ |
(I) $8|A|$ |
|
(B) $|A(adj\, A)|$ |
(II) $|A|^2$ |
|
(C) $|2A|$ |
(III) $\frac{1}{|A|}$ |
|
(D) $|A^{-1}|$ |
(IV) $|A|^3$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
|
List-I |
List-II |
|
(A) $|adj\,A|$ |
(II) $|A|^2$ |
|
(B) $|A(adj\, A)|$ |
(IV) $|A|^3$ |
|
(C) $|2A|$ |
(I) $8|A|$ |
|
(D) $|A^{-1}|$ |
(III) $\frac{1}{|A|}$ |
Given: $A$ is an invertible $3\times 3$ matrix.
(A) $|\text{adj }A|$
For $n\times n$: $|\text{adj }A| = |A|^{\,n-1}$. Here $n=3$, so $|\text{adj }A| = |A|^2$. Matches (II)
(B) $|A(\text{adj }A)|$
$A(\text{adj }A)=|A|I$, so determinant $=|A|^3$. Matches (IV)
(C) $|2A|$
$|kA|=k^n|A|$ with $n=3$, so $|2A|=2^3|A|=8|A|$. Matches (I)
(D) $|A^{-1}|$
$|A^{-1}|=\frac{1}{|A|}$. Matches (III)
The correct matching is: A–II, B–IV, C–I, D–III.